Compute the length of the curve f(x)=4√7(4−x2) over the interval 0≤x≤2. (Use decimal notation. Give your answer to three decimal places.)

Answer:

To compute the length of the curve f(x)=47(4−x2) over the interval 0≤x≤2, we need to use the formula for the arc length of a function:

L=∫ab1+(f′(x))2dx

where a and b are the endpoints of the interval. First, we need to find the derivative of f(x), which we can do by using the chain rule and the power rule:

f′(x)=4dxd7(4−x2)

f′(x)=427(4−x2)1dxd(7(4−x2))

f′(x)=427(4−x2)1(−14x)

f′(x)=−7(4−x2)28x

Next, we need to plug in f′(x) into the formula and simplify:

L=∫021+(−7(4−x2)28x)2dx

L=∫021+7(4−x2)784x2dx

L=∫027(4−x2)7(4−x2)+784x2dx

L=∫024−x228−21x2dx

Now, we need to evaluate the integral, which we can do by using a trigonometric substitution. Let x=2sinu, then dx=2cosudu and u=arcsin(x/2). The limits of integration change as follows:

x=0⟹u=0

x=2⟹u=2π

The integral becomes:

L=∫02π4−(2sinu)228−21(2sinu)2(2cosu)du

L=∫02π4−4sin2u28−84sin2u(2cosu)du

L=∫02π1−sin2u7−21sin2u(2cosu)du

L=∫02πcos2u7−21sin2u(2cosu)du

L=∫02π27−21sin2udu

Using a trigonometric identity, we can write:

L=∫02π4127−1221cos(2u)du

Using another trigonometric substitution, let v=2u, then dv=2du and u=v/2. The limits of integration change as follows:

u=0⟹v=0

u=2π⟹v=π

The integral becomes:

L=∫0π4127−1221cosv(21)dv

L=6∫0π